You are asked to design a cylindrical steel rod \({\bf{50}}{\bf{.0}}\;{\bf{cm}}\) long, with a circular cross-section that will conduct \({\bf{190}}{\bf{.0}}\;{{\bf{J}} \mathord{\left/{\vphantom {{\bf{J}} {\bf{s}}}} \right.\\} {\bf{s}}}\) from the furnace at \({\bf{400}}\;{\bf{^\circ C}}\) to a container of boiling water under \({\bf{1}}\) atmosphere. What must the rod’s diameter be?

The cross-sectional area of the rod is\({\rm{A}} = \pi {r^2}\).

The difference in temperature between the two ends:

\(\begin{array}{c}\Delta T = \left( {400 - 100} \right)\;{\rm{^\circ C}}\\ = {\rm{300}}\;{\rm{^\circ C}}\end{array}\).

The length of the steel rod:\(L = 50\;{\rm{cm}}\).

The thermal conductivity of the steel is:\({\kappa _S} = 50.2\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{m}} \cdot {\rm{K}}}}} \right.\\} {{\rm{m}} \cdot {\rm{K}}}}\).

The rate of heat transfer: \(190.0\;\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

The spontaneous transfer of heat, from a body which is at higher temperature to a body at \({\bf{a}}\) relatively lower temperature, when they are in contact is called thermal conduction. The transfer of heat will continue until the temperature of both bodies becomes equal. The thermal conductivity of an object is given as

\(\kappa {\bf{ = }}\frac{{\bf{L}}}{{{\bf{A\Delta T}}}}\frac{{{\bf{dQ}}}}{{{\bf{dt}}}}\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Using equation (1), the thermal conductivity of the bottom of the pot can be calculated as-

\(\kappa = \frac{L}{{A\Delta T}}\frac{{dQ}}{{dt}}\)

For the given values of the quantities, the equation becomes-

\(\begin{array}{l}50.2\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{m}} \cdot {\rm{K}}}}} \right.\\} {{\rm{m}} \cdot {\rm{K}}}} = \frac{{\left( {0.5\;{\rm{m}}} \right)}}{{\left( {\pi {r^2}} \right)\left( {300\;^\circ {\rm{C}}} \right)}} \times 190\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.\\} {\rm{s}}}\\{r^2} = 2 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{2}}}\\r = 4.5\;{\rm{cm}}\\{\rm{d}} = {\rm{9}}\;{\rm{cm}}\end{array}\)

The diameter of the rod is \(9\;{\rm{cm}}\).

The diameter of the cross-section of the rod is \(9\;{\rm{cm}}\).